Question

The sum to $n$ terms of $\left[\frac{1}{1.3}+\frac{2}{\mathrm{1.3.5}}+\frac{3}{\mathrm{1.3.5.7}}+\frac{4}{\mathrm{1.3.5.7.9}}+........\right]$ is

• A. $\frac{1}{2}\left[1+\frac{1}{\mathrm{1.3.5........}\left(2n+1\right)}\right]$
• B. $\frac{1}{2}\left[1-\frac{1}{\mathrm{2.4.6........2}n}\right]$
• C. $\frac{1}{2}\left[1-\frac{1}{\mathrm{1.3.5........}\left(2n+1\right)}\right]$
• D. None of these

Answer 1

Answer: (C)

$t_n=\frac{n}{\mathrm{1.3.5.}.........\left(2n+1\right)}$
$=\frac{1}{2}.\frac{\left(2n+1\right)-1}{\mathrm{1.3.5.}.........\left(2n+1\right)}$
$=\frac{1}{2}\left[\frac{1}{\mathrm{1.3.5......}\left(2n-1\right)}-\frac{1}{\mathrm{1.3.5.......}\left(2n+1\right)}\right]$
$=\frac{1}{2}\left(T_{n-1}-T_n\right)$
$\therefore 2t_n=T_{n-1}-T_n\dots \left(1\right)\left(where{T}_n=\frac{1}{\mathrm{1.3.5.......}\left(2n+1\right)}\right)$

$\therefore 2S_n=\sum _{n=2}^{n}2t_n+2t_1$
$\Rightarrow 2S_n-2t_1=(T_1-T_2)+(T_2-T_3)+........+(T_{n-1}-T_n)$
$\Rightarrow 2S_n-2t_1=T_1-T_n$
$$\begin{gathered} \Rightarrow 2S_n=2t_1+T_1-T_n \\ =2\times \frac{1}{3}+\frac{1}{1\times 3}-\frac{1}{1\times 3\times \mathrm{5.....}(2n+1)} \end{gathered}$$
$\Rightarrow S_n=\frac{1}{2}\left(1-\frac{1}{1\times 3\times \mathrm{5.......}(2n+1)}\right)$